TCP Sequence Number Wraparound Time at 2 Mbytes/sec

In TCP, the Sequence Number identifies the byte stream position. The sequence number field is 32 bits, so it conceptually wraps modulo $2^{32}$ bytes: after reaching $2^{32}-1$, the next increment goes back to $0$ 2.

Given:

• Sending rate: $2$ Mbytes/sec
• Initial sequence number value: starts with $7000$, i.e., we model it as $7000$
• We want the time until the sequence number returns to zero

We interpret “goes back to zero” as the sequence number field wrapping after the sender transmits enough data bytes such that the sequence number advances by $2^{32}$ bytes (because wrap is modulo $2^{32}$) 2.

We compute:

1. Remaining bytes until wrap from $7000$ to $0$ (inclusive wrap accounting): $$\text{bytes to wrap} = 2^{32} - 7000$$
2. Convert the rate $2$ Mbytes/sec to bytes/sec: Here we use $1$ Mbyte $= 10^6$ bytes (common in such problems unless stated otherwise): $$2\ \text{Mbytes/sec} = 2\times 10^6\ \text{bytes/sec}$$
3. Time: $$t=\frac{2^{32}-7000}{2\times 10^6}\ \text{sec}$$

Mermaid intuition for wraparound:

Now evaluate:

Convert to hours:

Answer:
[ \boxed{t \approx 2.15\times 10^3\ \text{seconds} \approx 35.8\ \text{minutes}} ] (Using $1$ Mbyte $=10^6$ bytes.)

Note on units: If your course instead uses $1$ MiB $=2^{20}$ bytes, the time will be slightly larger. The modulo logic $2^{32}-7000$ is the core part 2.